The infrared spectrum of chloroform (CHCl3) shows a band at 3034 cm-1 that is as

Question

The infrared spectrum of chloroform (CHCl3) shows a band at 3034 cm-1 that is assigned as the fundamental of the C-H stretch. The corresponding vibration of deuterated chloroform (CDCl3) appears at 2256 cm-1, so the ratio of the H and D frequencies is 3034/2256 = 1.345.
(a) Calculate the expected H to D ratio for a diatomic CH (or CD) molecule.
(b) Calculated the expected H to D ratio for a “diatomic molecule” H-CCl3 or D-CCl3, treating the CCl3 group as a single atom.
(c) Based on this information, what can you conclude about the amount that the Cl atoms move in this normal mode?

Explanation / Answer

(a) in case of diatomic CH (or CD) the C-H (or C-D) stretching frequency will be less. In CHCl3 and CDCl3, presence of an electronegative atom (-Cl atom) which has -I effect (inductive effect, electron withdrawing effect) results in the bond order to increase, hence increases the IR- stretching frequency. Isotopic effect (replacement of H by D) always reduce the bond order, hence decrease the IR frequency as D is heavier than H.

The expected C-H stretching frequency in C-H is 2942 cm-1, and that of C-D stretching frequency is 2200 cm-1, Hence H to D ratio is (2942/2200) = 1.33

(b) CCl3 has very strong electron withdrawing effect as compared to -Cl atom. Hence if we consider CCl3 group as a single atom then in CHCl3 and CDCl3, the C-H (or C-D) bond order will increase, results in higher IR-Stretching frequency. Expected IR-Stretching frequency of C-H in CHCl3 is 3250 cm-1 and C-D in CDCl3 is 2540 cm-1. Hence H to D ratio is (3250/2540) = 1.28

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