Q.30 In the 1 H NMR spectrum, which one of the following compounds will show a triplet?

The compound that shows a clear triplet in its 11H NMR spectrum is option (A), ethyl acetate (CH33COOCH22CH33).iit

Introduction

In many CSIR NET and GATE organic spectroscopy questions, you are asked which compound shows a triplet in its 11H NMR spectrum. Understanding spin–spin splitting and applying the $n+1$ rule lets you quickly recognize that ethyl acetate is the correct answer here. This article breaks down every option so you can see exactly how the splitting patterns arise.

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Option (A): Ethyl acetate – correct

Structure: CH33–CO–O–CH22–CH33 (an ethyl ester).

• The CH33CO– methyl group is adjacent to a carbonyl carbon that has no hydrogens, so it appears as a singlet (no neighboring protons to couple with).

• The –O–CH22– protons are adjacent to –CH33 (three equivalent hydrogens), so they are split into a quartet by 3 neighbors $(n+1=3+1=4)$.

• The terminal –CH33 next to CH22 has two neighboring hydrogens, so it is split into a triplet $(n+1=2+1=3)$; this classic ethyl pattern (triplet–quartet) is well documented for ethyl acetate.

Therefore, option (A) definitely shows a triplet signal in its 11H NMR spectrum.

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Option (B): Acetone – no triplet

Structure: CH33–CO–CH33.

• Both methyl groups are chemically equivalent and each is adjacent only to the carbonyl carbon, which carries no hydrogens.

• With zero vicinal neighbors, each methyl gives a singlet; because the two methyls are equivalent, the spectrum effectively shows one singlet corresponding to six protons.

Thus, there is no triplet for acetone; only a singlet is observed for its protons.

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Option (C): Aromatic methyl ester with two chloros – no simple triplet

Structure: a benzene ring bearing a CO22Me group and two chloro substituents (meta/para-disubstituted pattern).

• Aromatic protons on a substituted benzene are typically coupled to more than one type of neighboring proton, so they appear as complex multiplets rather than simple triplets.

• The methoxy group of an aryl ester (–CO22Me) appears as a singlet, because the three methoxy hydrogens are equivalent and usually have no vicinal hydrogens to couple with on oxygen.

Hence this compound shows multiplets and singlets, but no isolated triplet like in an ethyl group.

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Option (D): p‑Bromophenol – no triplet

Structure: para‑bromophenol, a benzene ring with Br and OH at para positions.

• Aromatic protons in a para‑disubstituted benzene typically give an AA′BB′ pattern, which appears as two sets of multiplets rather than clean triplets.

• The phenolic OH proton usually gives a broad singlet and may even disappear with exchange, again not a triplet.

Therefore, p‑bromophenol does not provide the distinct triplet requested in the question.

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Key takeaway: applying the $n+1$ rule

• A proton (or set of equivalent protons) split by n neighboring nonequivalent protons gives $n+1$ lines (doublet, triplet, quartet, etc.).

• Ethyl groups (–CH22–CH33) are classic: CH33 appears as a triplet (two neighbors on CH22), and CH22 appears as a quartet (three neighbors on CH33), as seen in ethyl acetate, option (A).