Question

10. Compound 6 reacts in the following Sn2 reaction to produce 7. 7 has the 13C NMR shown. What are 6 and 7. Br TTTT 60 40 200 100 om 80

Answer 1

Answer: "C" is the correct answer pa K coi 11 1. c) u Br how Explanation - We can eliminate "A", "B"and "D" because 0 The 130 nmr does not have any characteristic peak corresponding to aldehyde carbon (-&-H) in - the region of chemical shift values 190-200 ppm. 0 compound 6 in option "D" will not give the product mentioned in option "D". (wrong product) Dr Br+ noora * Bron ® Option "A" : the product bro By contains total it chemically non-equivalent carbon atome, hence 13C NMR should have 4 peaks in the spectrum. the peak around 30-50 PPM region corresponds to caybon attached to Bromine atom ((-Br) which is ing in the spectrum , hence option A was eliminated

verification of option "s" compound 6 reacts with given nucleophile in SN2 Reaction to give compound 7, which is correup 7 Br + por SN? E non + NaBr -Bro mechanisms the given alkoxine is a good nucleophile and attacks the carbon bearing bromine atom from backside then Bromide ion will be substituted a slow step, HH O BY --C--- "Br + Na Š. RDS Transition state B20 BT NaBr + now In the slow Step which is rare determining step two reactants are involved hence we call it as substitution Nucleophilic bi-molecular or SN?.

the compound 7 now) has three number of chemically nou equivalent carbons as shown below, hence we see three peaks in the given spectrum c-'arriba C- - be - - - PPM - 80 - 60 40 20 approrimate chemical shift (ppm) carbon environment N a w o The carbon attached to oxygen experinces more deshilding effect because of more electronegative ingen atom which attached to it, so the chemical value will be more than the other carbous in Same molecule.