46. Of the following, which is the best method to prepare t-butyl methyl ether?

Question

46. Of the following, which is the best method to prepare t-butyl methyl ether? a. CH OH+ Na; then (CH) CCI b. (CHs) COH +Na; then CHyl c. CHyBr+Mg; then (CH) COH d. CH OH +(CHs)hCOH with H SO4 and heat 47, The 'H NMR spectrum of a molecule containing the partial structure shown below would be expected to display which of the following? CH3 a. 1.7 (singlet, 3H), 5 4 (singlet, 1H) b, 1.7 (triplet, 3H), 5.4 (singlet, 1H) c, 1.7 (quartet, 3H), 5.4 (doublet, 1H) d, 1.7 (doublet, 3 H), 5.4 (quartet, 1 H) 48. What is the major product of the reaction below? Zn(Cu) a. 1 b. 2 c. 3 d. 4 49. What is the major organic product obtained from the following reaction? (CH3 Cui a.(E) 2-iodo-2-butene b. 1-iodo-2-methylpropane c. methylcyclopropane d. (E)-2-butene 50. Which of the following bonds has the most ionic character? a. C-Cu b. C-Zn c. C-Mg d. C-Li

Explanation / Answer

46. The correct answer is an option b.

Explanation: This is an example of Williamson's ether synthesis.

Here, both the options a and b are examples of Williamson's ether synthesis.

In the case of option a, CH3OH + Na gives CH3O-Na+, which acts as a strong base and causes an elimination (E2) reaction when reacts with (CH3)3CCl to give 2-butene, instead of MTBE.

Hence, the option b is a correct answer.

47. The correct answer is an option b.

48. The correct answer is an option c.

49. The correct answer is an option b.

50. The correct answer is an option d.

Explanation: The electronegativities of C = 2.5, Cu = 1.9, Zn = 1.7, Mg = 1.3, Li = 1.0

C-Li = 2.5-1.0 = 1.5

C-Mg = 2.5-1.3 = 1.2

C-Zn = 2.5-1.7 = 0.8

C-Cu = 2.5-1.9 = 0.6

Here, the electronegativity difference is more in the case of C-Li. Hence C-Li bond has the most ionic character.

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