Q.40 In the 1H NMR spectrum of 1–bromopropane (structure shown below), which of the
following statements are CORRECT?
CH2 CH2 CH3
cba
Br
(A) Protons ‘a’ resonate upfield to protons ‘c’
(B) Protons ‘b’ resonate downfield to protons ‘c’
(C) There are two triplets and one quartet in the spectrum
(D) Protons ‘a’ appear as triplet with high chemical shift in comparison to protons ‘c’

Protons b resonate downfield to protons c, and protons a appear as a triplet at the highest chemical shift compared with protons c; the spectrum shows two triplets and one complex multiplet (not a quartet).​
Therefore, options B and D are correct, whereas A and C are incorrect.​​

Introduction

The 1H NMR spectrum of 1‑bromopropane (Br–CH2–CH2–CH3) is a classic exam problem that tests understanding of chemical shift trends and spin–spin splitting in simple alkyl halides.​
In this molecule there are three nonequivalent sets of protons, labelled a (Br–CH2–), b (central –CH2–), and c (terminal –CH3), each showing distinct positions and multiplicities in the spectrum.​

Assigning chemical shifts and multiplicities

• Proton set a (Br–CH2–) is strongly deshielded by the electronegative bromine and appears farthest downfield, around 3.3–3.5 ppm as a triplet, because it couples with the two equivalent b‑protons.​

• Proton set b (middle –CH2–) lies between two carbon atoms, one bearing Br and one bearing CH3, and therefore appears at intermediate shift (about 1.6–2.1 ppm) as a complex multiplet (approximately sextet) due to coupling with both a (2 H) and c (3 H).​

• Proton set c (terminal –CH3) is farthest from bromine, most shielded, and appears upfield near 0.9–1.1 ppm as a triplet by coupling with the two equivalent b‑protons.​

From these data, the order of chemical shift is: a (most downfield) > b > c (most upfield).​

Detailed analysis of each option

Option (A): “Protons ‘a’ resonate upfield to protons ‘c’”

• Upfield means at lower ppm (more shielded), while downfield means at higher ppm (more deshielded).​

• Because the a‑protons are directly attached to the bromine‑bearing carbon, they are more deshielded and thus resonate downfield of c, not upfield; therefore, statement (A) is incorrect.​

Option (B): “Protons ‘b’ resonate downfield to protons ‘c’”

• The central b‑CH2 group is closer to bromine than the terminal c‑CH3 and thus experiences a stronger electron‑withdrawing effect.​

• Consequently, b appears at higher ppm (more downfield) than c, so statement (B) is correct.​

Option (C): “There are two triplets and one quartet in the spectrum”

• In the actual 1H NMR of 1‑bromopropane, a and c each appear as triplets, but the b‑protons appear as a complex multiplet (approximately sextet), not a quartet, due to coupling with five neighboring protons (2 from a and 3 from c).​

• Therefore, the pattern is “two triplets and one multiplet,” not “two triplets and one quartet,” making statement (C) incorrect.​

Option (D): “Protons ‘a’ appear as triplet with high chemical shift in comparison to protons ‘c’”

• As noted above, a‑protons couple only with the two equivalent b‑protons, giving an $n+1=3$ line pattern, i.e., a triplet.​

• They also resonate at significantly higher ppm (downfield) than c because of direct attachment to the brominated carbon, so they have the higher chemical shift relative to c; thus, statement (D) is correct.​

Summary table of proton environments in 1‑bromopropane

In competitive examinations, remember that in 1‑bromopropane, the correct statements about the 1H NMR spectrum are B and D, guided by both electronegativity effects on chemical shift and the $n+1$ rule for splitting.​