---
title: "Reaction Mechanisms: Curly Arrows"
book: "School Chemistry — Grades 1 to 12"
subject: chemistry
language: en
chapter: 40
exercises: 15
source: https://one-course.com/books/chemistry/1/en/chapter/40-reaction-mechanisms-curly-arrows
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 40 — Reaction Mechanisms: Curly Arrows

A [balanced equation](https://one-course.com/books/chemistry/1/en/chapter/13-conservation-of-mass-and-balanced-equations#def-g8-balanced-equations-equation) is like the first and last photographs of a film: it shows the [reactants](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) before and the products after, and nothing in between. Yet [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) do not simply swap [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). Bonds break and form one at a time, as pairs of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) move from one place to another, through short-lived species that never appear in the equation. Chemists draw these movements with curved arrows. With a few rules, these arrows explain why a reaction happens where it does, and predict reactions never seen before.

**You already know.**

[Electronegativity](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-electronegativity), [polar bonds](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-polar-bond) and [partial charges](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-polar-bond) ([Chapter 30](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#ch-g11-polarity-and-cohesion)). [Lewis structures](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lewis-structure): [bonding pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) and [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) ([Chapter 24](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#ch-g10-lewis-and-shape)). [Functional groups](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-functional-group) and families ([Chapter 33](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#ch-g11-functional-groups)).

## 40.1 Electron-rich and electron-poor sites

**Definition 40.1 (Electron-donor and electron-acceptor sites).**

An *electron-donor site* of a [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) or an [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is a place rich in [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), able to give a pair to form a new bond: a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair), a [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond), an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) carrying a negative charge or a [partial charge](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-polar-bond) $\delta^-$. An *electron-acceptor site* is a place poor in [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), able to receive a pair: an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) carrying a positive charge or a [partial charge](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-polar-bond) $\delta^+$.

**Example 40.2 (Sites in two molecules).**

In bromoethane, $\ce{CH3-CH2-Br}$, bromine is more electronegative than carbon: the carbon bonded to it carries $\delta^+$ (acceptor site), the bromine $\delta^-$ and three [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair). In propanone, the oxygen of the $\ce{C=O}$ group carries $\delta^-$ and two [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) (donor sites), its carbon $\delta^+$ (acceptor site). The hydroxide [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{OH-}$, with its negative charge and three [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair), is a strong donor.

![Electron-poor carbon atoms (+, acceptor sites) next to electron-rich, electronegative atoms (-, with lone pairs: donor sites).](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-curly-arrows/fig-e394b8fe91aa.svg)

*Electron-poor carbon [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) ($\delta^+$, acceptor sites) next to electron-rich, electronegative [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) ($\delta^-$, with [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair): donor sites).*

## 40.2 Curly arrows

**Definition 40.3 (Curly arrow).**

A *curly arrow* shows the movement of a pair of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) during a step of a reaction. It starts from the pair that moves, a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) or a bond, and ends where the pair goes: on an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), where it forms a new bond, or on an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) that takes the pair of a bond that breaks.

**Method 40.4 (Drawing curly arrows).**

1. Find the donor site and the acceptor site.
2. Draw an arrow from the donor pair (a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) or a bond) to the acceptor [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) : a new bond forms there.
3. If the acceptor [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) would then have too many [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) (more than an octet, or more than two for hydrogen), draw a second arrow breaking one of its bonds, the pair going to the more electronegative [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) .
4. Check the charges after the step: the [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) that gave a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) becomes more positive by one, the [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) that received a pair as a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) more negative by one; the total charge does not change.

**Proposition 40.5 (Charge is conserved in each step).**

The total electric charge of the species is the same before and after each step of a mechanism.

**Proof.** A [curly arrow](#def-g12-curly-arrows-curly-arrow) only moves [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) that are already there; no [electron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) is created or destroyed, and the nuclei do not change. ∎

**Example 40.6 (A first arrow).**

A hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{H+}$ meets a hydroxide [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{OH-}$. A [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of the oxygen forms a bond with the hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), which has no [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus): $\ce{H+ + OH- -> H2O}$. One arrow, from the [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of the oxygen to the $\ce{H+}$; the charges $+1$ and $-1$ become 0 in the water [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule).

![The simplest step: a lone pair of the hydroxide ion moves to form the new O-H bond with the hydrogen ion.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-curly-arrows/fig-955ecc93764f.svg)

*The simplest step: a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of the hydroxide [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) moves to form the new $\ce{O-H}$ bond with the hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion).*

## 40.3 Elementary steps and intermediates

**Definition 40.7 (Mechanism, elementary step, intermediate).**

The *reaction mechanism* of a reaction is the series of steps by which the [reactants](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) become the products. Each *elementary step* is a single event in which a few pairs of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) move at once. A species formed in one step and used up in a later one is a *reaction intermediate*: it does not appear in the overall equation.

**Example 40.8 (A carbocation intermediate).**

When hydrogen bromide adds to ethene, the reaction takes two steps. In the first, the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond) of ethene gives a pair to the hydrogen of $\ce{HBr}$, and the $\ce{H-Br}$ pair goes to bromine, which leaves as $\ce{Br-}$; the carbon that lost its share of the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond) is left with only six [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) and a positive charge: a carbocation, $\ce{CH3-CH2+}$. In the second step, a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of $\ce{Br-}$ forms a bond with that carbon. The carbocation is an intermediate.

## 40.4 Three categories of reaction

**Definition 40.9 (Substitution, addition, elimination).**

In a *substitution*, an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) or group of a [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) is replaced by another. In an *addition*, two [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) join into one, a multiple bond becoming a single one. In an *elimination*, a small [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) (often water) is removed from a [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule), a single bond becoming a multiple one.

![Three mechanisms. Orange arrows: a donor pair forms a new bond. Blue arrows: a bond breaks, its pair going to the more electronegative atom. In the elimination the hydrogen ion is given back at the end.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-curly-arrows/fig-2e128f5f4357.svg)

*Three mechanisms. Orange arrows: a donor pair forms a new bond. Blue arrows: a bond breaks, its pair going to the more electronegative [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). In the [elimination](#def-g12-curly-arrows-categories) the hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is given back at the end.*

**Remark 40.10 (Changing the skeleton or the group).**

Some reactions change only the [functional group](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-functional-group) and keep the carbon skeleton: the [substitution](#def-g12-curly-arrows-categories) of bromoethane turns a [halogenoalkane](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-halogenoalkane) into an [alcohol](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-alcohol), two carbons before, two after. Others build or cut the skeleton: making a new carbon–carbon bond lengthens a chain, the key step in making a large [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) from small ones. Planning a [synthesis](https://one-course.com/books/chemistry/1/en/chapter/28-synthesis-yield-and-purity#def-g10-synthesis-yield-synthesis) means choosing reactions of both kinds, in the right order.

## 40.5 Exercises

**Exercise 40.1 ★.**

In each species, name a donor site and, if there is one, an acceptor site: $\ce{OH-}$; $\ce{H+}$; $\ce{CH2=CH2}$; $\ce{CH3-Cl}$; $\ce{H2O}$.

**Solution of Exercise 40.1.**

$\ce{OH-}$: donor ([lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair), negative charge); no acceptor site. $\ce{H+}$: acceptor (no [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) at all). $\ce{CH2=CH2}$: donor (the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond)). $\ce{CH3-Cl}$: donor ([lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of $\ce{Cl}$), acceptor (the carbon, $\delta^+$). $\ce{H2O}$: donor ([lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of $\ce{O}$); its hydrogen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom), $\delta^+$, are weak acceptor sites.

**Exercise 40.2 ★.**

[Substitution](#def-g12-curly-arrows-categories), [addition](#def-g12-curly-arrows-categories) or [elimination](#def-g12-curly-arrows-categories)? (a) $\ce{CH3-CH2-Cl + OH- -> CH3-CH2-OH + Cl-}$; (b) $\ce{CH2=CH2 + H2O -> CH3-CH2-OH}$; (c) $\ce{CH3-CH2-OH -> CH2=CH2 + H2O}$; (d) $\ce{CH2=CH2 + Br2 -> CH2Br-CH2Br}$.

**Solution of Exercise 40.2.**

(a) [substitution](#def-g12-curly-arrows-categories); (b) [addition](#def-g12-curly-arrows-categories); (c) [elimination](#def-g12-curly-arrows-categories); (d) [addition](#def-g12-curly-arrows-categories).

**Exercise 40.3 ★.**

Where does a [curly arrow](#def-g12-curly-arrows-curly-arrow) start, and where does it end? What does a [curly arrow](#def-g12-curly-arrows-curly-arrow) from a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of an oxygen [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) to a carbon [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) mean?

**Solution of Exercise 40.3.**

It starts from a pair of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) (a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) or a bond) and ends on the [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) where the pair goes. From a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of oxygen to a carbon: the pair becomes a new $\ce{O-C}$ bond.

**Exercise 40.4 ★.**

In the reaction $\ce{H+ + NH3 -> NH4+}$, which species gives the pair? Draw the arrow. What is the charge of the product?

**Solution of Exercise 40.4.**

Ammonia gives the [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of its nitrogen; the arrow goes from that [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) to $\ce{H+}$. The product $\ce{NH4+}$ carries the charge $+1$.

**Exercise 40.5 ★.**

What is a [reaction intermediate](#def-g12-curly-arrows-mechanism)? Name the intermediate of the [addition](#def-g12-curly-arrows-categories) of $\ce{HBr}$ to ethene.

**Solution of Exercise 40.5.**

A species formed in one step and used up in a later one; it is not in the overall equation. Here the carbocation $\ce{CH3-CH2+}$.

**Exercise 40.6 ★★.**

Draw the [curly arrows](#def-g12-curly-arrows-curly-arrow) of the [substitution](#def-g12-curly-arrows-categories) $\ce{CH3-Br + OH- -> CH3-OH + Br-}$ (one step), and check the charges.

**Solution of Exercise 40.6.**

One arrow from a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of the oxygen of $\ce{OH-}$ to the carbon; one arrow from the $\ce{C-Br}$ bond to the bromine. Charges: $-1$ on the left ($\ce{OH-}$); on the right, methanol is neutral and $\ce{Br-}$ carries $-1$: the total is $-1$ on both sides.

**Exercise 40.7 ★★.**

In the first step of the [addition](#def-g12-curly-arrows-categories) of $\ce{HBr}$ to ethene, why does the $\ce{H-Br}$ pair go to bromine rather than to hydrogen? What charge does each product of the step carry?

**Solution of Exercise 40.7.**

Bromine is more electronegative than hydrogen: it takes the pair and leaves as $\ce{Br-}$ ($-1$); the [hydrogen bonds](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-hydrogen-bond) to a carbon, and the other carbon is left with a positive charge: $\ce{CH3-CH2+}$ ($+1$).

**Exercise 40.8 ★★.**

In the dehydration of ethanol, write the three steps and say, for each, which species is the donor and which the acceptor. Why is the hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) not written in the overall equation?

**Solution of Exercise 40.8.**

1. $\ce{CH3-CH2-OH + H+ -> CH3-CH2-OH2+}$: donor, a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of the oxygen; acceptor, $\ce{H+}$. 2. $\ce{CH3-CH2-OH2+ -> CH3-CH2+ + H2O}$: the $\ce{C-O}$ pair goes to the positive oxygen (acceptor), which leaves with it as water. 3. $\ce{CH3-CH2+ -> CH2=CH2 + H+}$: the pair of a $\ce{C-H}$ bond (donor) moves to the positive carbon (acceptor) to form the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond), and $\ce{H+}$ leaves. The hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) used in step 1 is given back in step 3: it appears on both sides and cancels.

**Exercise 40.9 ★★.**

The step $\ce{CH3-CH2+ + H2O -> CH3-CH2-OH2+}$ follows the [addition](#def-g12-curly-arrows-categories) of a hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) to ethene in water. Draw the arrow. What is formed after a further loss of $\ce{H+}$? Write the overall equation and give its category.

**Solution of Exercise 40.9.**

An arrow from a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of the oxygen of water to the positive carbon. After loss of $\ce{H+}$, ethanol $\ce{CH3-CH2-OH}$. Overall: $\ce{CH2=CH2 + H2O -> CH3-CH2-OH}$, an [addition](#def-g12-curly-arrows-categories).

**Exercise 40.10 ★★.**

Read the mechanism of the [substitution](#def-g12-curly-arrows-categories) of bromoethane: how many pairs move? Which bond forms, which breaks? Is there an intermediate?

**Solution of Exercise 40.10.**

Two pairs move, in one step: the $\ce{C-O}$ bond forms, the $\ce{C-Br}$ bond breaks. No intermediate.

**Exercise 40.11 ★★.**

Chloromethane reacts with the hydroxide [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) more slowly than bromomethane. Using electronegativities (C 2.55, Cl 3.16, Br 2.96), explain which carbon is the more $\delta^+$. Does that explain the speeds? (Think also of how easily the leaving [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) takes the pair.)

**Solution of Exercise 40.11.**

Chlorine is more electronegative ($3.16 - 2.55 = 0.61$ against $2.96 - 2.55 = 0.41$): the carbon of chloromethane is the more $\delta^+$, which alone would make it react faster. It does not: the speed also depends on how easily the [halogen](https://one-course.com/books/chemistry/1/en/chapter/23-electron-shells-and-the-periodic-table#def-g10-electron-shells-family) leaves with the pair, and the larger bromine leaves more easily.

**Exercise 40.12 ★★★.**

Hydrogen chloride adds to propene, $\ce{CH2=CH-CH3}$. In the first step the hydrogen [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) can bond to either carbon of the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond), giving two possible carbocations. Draw both, the two possible products, and name them. (Which one forms most is studied in the Year 1 volume.)

**Solution of Exercise 40.12.**

If the [hydrogen bonds](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-hydrogen-bond) to carbon 1: $\ce{CH3-CH+-CH3}$, which gives 2-chloropropane $\ce{CH3-CHCl-CH3}$. If it bonds to carbon 2: $\ce{+CH2-CH2-CH3}$, which gives 1-chloropropane $\ce{CH2Cl-CH2-CH3}$.

**Exercise 40.13 ★★★.**

A student draws the first step of the [substitution](#def-g12-curly-arrows-categories) of bromoethane with an arrow from the bromine [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) to the oxygen of $\ce{OH-}$. Explain the two mistakes.

**Solution of Exercise 40.13.**

The arrow goes the wrong way: it must start from the donor, a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of $\ce{OH-}$, and end on the acceptor carbon. And the new bond is between oxygen and carbon, not oxygen and bromine: bromine leaves, taking the pair of the $\ce{C-Br}$ bond.

**Exercise 40.14 ★★★.**

In the formation of an [ester](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-carboxylic-acid) from a [carboxylic acid](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-carboxylic-acid) and an [alcohol](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-alcohol), the oxygen of the [alcohol](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-alcohol) attacks the carbon of the $\ce{C=O}$ group. Using [partial charges](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-polar-bond), explain why that carbon is an acceptor site and that oxygen a donor site. Which small [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) is lost overall, and what category is the whole reaction?

**Solution of Exercise 40.14.**

The carbon of the $\ce{C=O}$ group of an acid is bonded to two electronegative oxygen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom): strongly $\delta^+$, an acceptor. The oxygen of the [alcohol](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-alcohol) carries $\delta^-$ and two [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair): a donor. Water is lost overall; the $\ce{-OH}$ of the acid is replaced by the $\ce{-O-R}$ of the [alcohol](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-alcohol): a [substitution](#def-g12-curly-arrows-categories).

**Exercise 40.15 ★★★.**

Ethene reacts with bromine $\ce{Br2}$, a [non-polar molecule](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-polar-molecule). As a $\ce{Br2}$ [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) approaches the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond), the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) of $\ce{Br-Br}$ are pushed away from the near bromine [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom). Explain why this creates an acceptor site, and propose the first step of the mechanism.

**Solution of Exercise 40.15.**

Pushed away by the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) of the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond), the pair of the $\ce{Br-Br}$ bond moves towards the far bromine: the near bromine becomes $\delta^+$, an acceptor site. First step: an arrow from the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond) to the near bromine, and one from the $\ce{Br-Br}$ bond to the far bromine, which leaves as $\ce{Br-}$; a positive charge is left on the other carbon of the former [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond).

## 40.6 Problem: Making Bromoethane

**Problem 40.1.**

Weekend problem — from ten grams of ethene: how much bromoethane?

Bromoethane, a starting material for many syntheses, is made by adding hydrogen bromide to ethene: $\ce{CH2=CH2 + HBr -> CH3-CH2-Br}$. A chemist uses $10.0\,\mathrm{g}$ of ethene and an excess of hydrogen bromide; the [yield](https://one-course.com/books/chemistry/1/en/chapter/28-synthesis-yield-and-purity#def-g10-synthesis-yield-yield) is $75\,\%$.

**Part I — The sites.**

1. Draw the [Lewis structures](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lewis-structure) of ethene and of hydrogen bromide.
2. Where is the donor site of ethene? Why?
3. Compute the difference of [electronegativity](https://one-course.com/books/chemistry/1/en/chapter/30-electronegativity-polarity-and-intermolecular-forces#def-g11-polarity-and-cohesion-electronegativity) of the $\ce{H-Br}$ bond (H 2.20, Br 2.96). Which [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) carries $\delta^+$ ?
4. Which [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) of $\ce{HBr}$ is the acceptor site?

**Part II — The mechanism.**

5. Draw the first step with its two [curly arrows](#def-g12-curly-arrows-curly-arrow) .
6. Give the two species formed and their charges.
7. Check the conservation of charge in the first step.
8. Draw the second step with its [curly arrow](#def-g12-curly-arrows-curly-arrow) .
9. How many [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) surround the positive carbon of the intermediate? Why is it so reactive?

**Part III — What kind of reaction?**

10. What is the intermediate of the mechanism?
11. Why does it not appear in the overall equation?
12. To which category does the reaction belong?
13. Does the reaction change the carbon skeleton, or only the [functional group](https://one-course.com/books/chemistry/1/en/chapter/33-functional-groups-and-families#def-g11-functional-groups-functional-group) ?

**Part IV — How much product?**

14. Compute the molar masses of ethene and of bromoethane.
15. Compute the amount of ethene.
16. Fill the [progress table](https://one-course.com/books/chemistry/1/en/chapter/27-the-reaction-progress-table#def-g10-reaction-progress-table-extent) . Which [reactant](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) is limiting?
17. Compute the maximum mass of bromoethane.
18. Compute the mass obtained with a [yield](https://one-course.com/books/chemistry/1/en/chapter/28-synthesis-yield-and-purity#def-g10-synthesis-yield-yield) of $75\,\%$ .
19. State the final answer: what mass of bromoethane does the chemist obtain?

**Solution of Problem 40.1.**

**1.** Ethene: $\ce{H2C=CH2}$, a [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond) and four $\ce{C-H}$. Hydrogen bromide: $\ce{H-Br}$ with three [lone pairs](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) on bromine.

**2.** The [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond): its second pair is a region rich in [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**3.** $2.96 - 2.20 = 0.76$: $\ce{H}$ carries $\delta^+$.

**4.** The hydrogen [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**5.** An arrow from the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond) to the hydrogen of $\ce{HBr}$; an arrow from the $\ce{H-Br}$ bond to the bromine.

**6.** The carbocation $\ce{CH3-CH2+}$ ($+1$) and $\ce{Br-}$ ($-1$).

**7.** Before: $0 + 0 = 0$; after: $+1 - 1 = 0$.

**8.** An arrow from a [lone pair](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-lone-pair) of $\ce{Br-}$ to the positive carbon.

**9.** Six (three bonds): one pair short of an octet, so it accepts a pair from any donor nearby.

**10.** The carbocation $\ce{CH3-CH2+}$.

**11.** It is formed in the first step and used up in the second.

**12.** An [addition](#def-g12-curly-arrows-categories).

**13.** Only the group: the chain of two carbons is kept, and the [double bond](https://one-course.com/books/chemistry/1/en/chapter/24-lewis-structures-and-the-shape-of-molecules#def-g10-lewis-and-shape-multiple-bond) becomes a single bond carrying $\ce{H}$ and $\ce{Br}$.

**14.** $28.0\,\mathrm{g}/\mathrm{mol}$; $108.9\,\mathrm{g}/\mathrm{mol}$.

**15.** $10.0 / 28.0 = 0.357\,\mathrm{mol}$.

**16.** Ethene: $0.357 - x$; $\ce{HBr}$: in excess; bromoethane: $x$. Ethene is limiting: $x_{\max} = 0.357\,\mathrm{mol}$.

**17.** $0.357 \times 108.9 = 38.9\,\mathrm{g}$.

**18.** $0.75 \times 38.9 = 29.2\,\mathrm{g}$.

**19.** About $29\,\mathrm{g}$ of bromoethane.