Question

According to the n + 1 rule, what would be the multiplicities of the signals in the proton NMR spectrum of 1,1‐ dibromoethane?

• A. Triplet and doublet
• B. Quartet and doublet
• C. Triplet and quartet
• D. Triplet and triplet

Answer 1

The Correct Option is B

The question asks about the multiplicities of proton NMR signals in the compound 1,1-dibromoethane using the n + 1 rule. To solve this, let's first understand the structure of 1,1-dibromoethane and how the NMR rule applies.

1,1-dibromoethane has the structure CH₂Br₂-CH₃. There are two types of hydrogen environments in this molecule:

• Methylene protons (CH₂): These are the two protons attached to the same carbon as the bromine atoms.
• Methyl protons (CH₃): These are the three protons attached to the terminal carbon atom.

The n + 1 rule helps to determine the multiplicity (splitting pattern) of a signal in NMR spectroscopy. According to the rule:

• The multiplicity of an NMR signal is determined by the number of neighboring (adjacent) hydrogen atoms (n) plus one.

Applying the n + 1 rule to the 1,1-dibromoethane:

1. Methyl protons (CH₃): There are no adjacent protons connected to the CH₂ group since it is directly attached to bromine. However, these three protons will sense the two adjacent CH₂ protons, giving a triplet signal. According to n + 1: n = 2, so n + 1 = 3 (triplet).
2. Methylene protons (CH₂): These two protons are adjacent to three methyl protons (CH₃). Thus, it will appear as a quartet signal. According to n + 1: n = 3, so n + 1 = 4 (quartet).

Therefore, the multiplicities of the signals in the proton NMR spectrum of 1,1-dibromoethane are quartet and doublet, respectively.

Hence, the correct answer is: Quartet and doublet.