For #14. and #15. Consider this reaction: (CH_3)_3CBr + H_20 rightarrow (CH_3)_3
ID: 996386 • Letter: F
Question
For #14. and #15. Consider this reaction: (CH_3)_3CBr + H_20 rightarrow (CH_3)_3COH + HBr What effect does doubling the concentration of water have on the rate? doubles the rate halves the rate quadruples the rate has no effect on the rate What effect does doubling the concentration of (CH_3)_3CBr have on the rate doubles the rate halves the rate quadruples the rate has no effect on the rate In the term Prime E2 Prime, what does the E indicate and what does the 2 indicate? How could you distinguish between these two compounds in the IR? What distinguishing peak/bands would you see for which compound? x = y = CH_2=CHC H_2C H_2OC H_3Explanation / Answer
14) The reaction will follow SN1 mechanism and hence there will be no effect of concentration of nucleophile (H2O)
Answer: has not effect on rate
15) The rate of reaction is directly proportional to concentration of substrate [(CH3)3CBr] so the rate of reaction will get double
Anser: doubles the rate
16) In the term E2
E : Elimination
2 : Bimolecular
[Which means bimolecular elimination , second order reaction ]
17) In compound X we will observe a peak near 1700 cm^-1 due to C=O bond
And absence of peak near 3000-3100 cm-1
In compound Y we will observe a peak near 3000- 3100 cm-1 due to C=C bond stretching
And a peak near 1050-1100 due to C-O bond