Question

a. CH1003 HIT-NO-974 IR-NIDA-08701 : CCL4 SOLUTION SCORE- SBS-NO- SOBS-NO- CHOS TERBOTTEN O RI1 1478 SBI 1913 1450 65 1297 10827 70 1446 63 1106 B 624 04 1418 52 1154 21 643 72 9 12 10 4 50001 108 1042 1723 4 1H NMR signals: 1.2 (triplet, 3H) 2.1 (singlet, 3H) 3.4 (singlet, 2H) 4.2 (quartet, 2H)

Answer 1

Solution:

In the given IR spectrum, asymmetric stretch is observed at 1745 and symmetric stretch is at 1723 for carbonyl groups, both are separated by 22 cm-1.

C-H stretch is found at 2984 cm-1.

Two C-O stretch bands are observed around 1000 cm-1.

Therefore, the given compound contains diketone functional group.

From the NMR data, triplet and quartet are adjacent to each other. So, it may be ethyl substituent.

singlet, 3H at 2.1 ppm comes from methyl substituent at one of the end of the diketone while ethyl group is another end.

Singlet at 3.4 ppm shows that it is present between two carbonyl groups.

Hence, the given compound is ethyl acetoacetate.