The infrared spectrum of a colourless liquid, compound Z, shows two separate medium-intensity absorptions at 3300 cm^-1 and 3380 cm^-1, together with a C–H absorption in the range 2850–2960 cm^-1. There is no absorption anywhere in the range 1650–1750 cm^-1, and no absorption in…

Answer: ((a)) Compound Z is a primary amine, -NH2: two separate absorptions in the N–H/O–H region (3300 and 3380 cm⁻¹) match the symmetric and asymmetric N–H stretches of an -NH2 group, whereas a secondary amine gives only one N–H absorption, and an alcohol gives only one (broad) O–H absorpti… ((b)) A primary amide would show a C=O absorption in the 1650–1750 cm⁻¹ range in addition to its N–H absorptions; since no absorption appears anywhere in this range for compound Z, it cannot contain a carbonyl group, so it cannot be an amide, despite having N–H bonds present. ((c)) A single N–H absorption from a secondary amine and a single O–H absorption from an alcohol can both fall within the same 3300–3600 cm⁻¹ region, so the peak position/count alone cannot distinguish them; further evidence such as the presence or absence of a C–N absorption, or a si…. Part (a): Identifying compound Z from the N–H region The data table shows that the number of separate absorptions in the 3200–3600 cm^-1 region is diagnostic: - A primary amine, -NH2, has two N–H bonds that vibrate independently (symmetric…

AS Chemistry 9701: Distinguishing an Amine from an Alcohol Using N-H and O-H Infrared Absorptions

Bond	Functional group	Wavenumber / cm⁻¹	Typical peak shape
N–H	primary amine, $-\text{NH}_2$	3300–3500	two absorptions (symmetric and asymmetric N–H stretch)
N–H	secondary amine, $-\text{NH}-$	3300–3500	one absorption
O–H	alcohol (hydrogen bonded)	3200–3600	one broad absorption
C=O	carbonyl (aldehyde, ketone, acid, ester, amide)	1650–1750	one strong, sharp absorption
C–H	alkyl	2850–2960	one or more absorptions

Bond

Functional group

Wavenumber / cm⁻¹

Typical peak shape

N–H

primary amine,

-\text{NH}_2

3300–3500

two absorptions (symmetric and asymmetric N–H stretch)

N–H

secondary amine,

-\text{NH}-

3300–3500

one absorption

O–H

alcohol (hydrogen bonded)

3200–3600

one broad absorption

C=O

carbonyl (aldehyde, ketone, acid, ester, amide)

1650–1750

one strong, sharp absorption

C–H

alkyl

2850–2960

one or more absorptions

Part (a): Identifying compound Z from the N–H region

The data table shows that the number of separate absorptions in the $3200$ – $3600\ \text{cm}^{-1}$ region is diagnostic:

A primary amine, $-\text{NH}_2$ , has two N–H bonds that vibrate independently (symmetric and asymmetric stretching), giving two separate absorptions in this region.
A secondary amine, $-\text{NH}-$ , has only one N–H bond, giving one absorption.
An alcohol O–H bond also gives only one (typically broader) absorption.

Compound Z shows two separate medium-intensity absorptions, at $3300\ \text{cm}^{-1}$ and $3380\ \text{cm}^{-1}$ . This matches the symmetric and asymmetric N–H stretches of a primary amine, and rules out both a secondary amine and an alcohol, each of which would give only a single absorption here.

Compound Z contains a primary amine group, $-\text{NH}_2$ .

Part (b): Ruling out a primary amide

A primary amide, $\text{R-CONH}_2$ , also contains N–H bonds and so would also absorb in the $3300$ – $3500\ \text{cm}^{-1}$ region, but it additionally contains a carbonyl, C=O, group, which would produce a strong, sharp absorption somewhere in the $1650$ – $1750\ \text{cm}^{-1}$ range.

Since compound Z shows no absorption anywhere in $1650$ – $1750\ \text{cm}^{-1}$ , it cannot contain a C=O bond, and so it cannot be an amide, despite the N–H absorptions being consistent with one. The N–H absorptions alone must instead come from an amine.

Part (c): Why one N–H/O–H peak is ambiguous

A secondary amine gives exactly one N–H absorption in $3300$ – $3500\ \text{cm}^{-1}$ , and an alcohol gives exactly one O–H absorption in the overlapping range $3200$ – $3600\ \text{cm}^{-1}$ . Since these two ranges overlap and both functional groups give only a single peak, seeing just one absorption there is consistent with either possibility, the peak count that worked to identify a primary amine in part (a) cannot, on its own, distinguish a secondary amine from an alcohol.

To decide between them, further evidence is needed: for example, checking for a C–N absorption (distinct from the C–O absorption an alcohol would show), or carrying out a simple chemical test that responds differently to an amine and an alcohol.

Final answers

(a) Primary amine ( $-\text{NH}_2$ ); two N–H absorptions (rather than one) distinguish it from a secondary amine and from an alcohol.
(b) No C=O absorption is present (1650–1750 cm⁻¹ is empty), so compound Z cannot be an amide even though it has N–H bonds.
(c) A secondary amine and an alcohol can both give a single peak in overlapping wavenumber ranges, so further evidence (e.g. a C–N absorption or a chemical test) is needed to distinguish them.