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JEE Advanced levelSingle correct optionDifficulty 3 of 3si090

A compound shows molecular ion peaks at m/z 112 and 114 with intensity ratio about $3 : 1$ and contains 64.0% C, 4.4% H and 31.6% Cl by mass (C 12, H 1, Cl 35.5). Its molecular formula and degree of unsaturation are

A $C X_{6} H X_{5} Cl$ and 3
B $C X_{6} H X_{5} Cl$ and 4
C $C X_{5} H X_{9} Cl$ and 1
D $C X_{6} H X_{4} Cl X_{2}$ and 4

Answer: B

A

3 : 1

M : M+2 pattern shows one Cl atom, so M(35Cl) = 112 leaves 77 for C and H. The percentages give

C X_{6} H X_{5}

(72 + 5 = 77), so the formula is

C X_{6} H X_{5} Cl

. Degree of unsaturation

= (14 - 5 - 1) /2 = 4

.

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