Question. Treatment of (CH₃)₂ CHCH₂CH(OH)CH₂CH₃ with TsOH affords two products (M and N) with the molecular formula C₆H₁₂ . The 1 H NMR spectra of M and N are given below. Propose structures for M and N, and draw a mechanism to explain their formation.

Tosyl chloride would activate the alcoholic group because of its electron-withdrawing nature. It acts as a protecting group and converts alcohols to corresponding alkenes.

When 5-Methylhex-3-ol is treated with tosyl chloride, the -OH group abstracts a proton to undergo dehydration.

The carbocation would undergo a 1,2-hydride shift followed by abstraction of an alpha hydrogen by the tosyl anion to give a mixture of alkenes.

a. Consider the NMR spectra of M:

• The two singlets at 1.6 attribute to 3H due to the adjacent carbon with no hydrogen atoms.
• The triplet at 5.1 attributes to 1H due to the adjacent 2H protons.
• The multiplet at 2.0 attributes to 2H due to the adjacent 1H and 3H protons.
• The triplet at 0.95 attributes to 3H due to the adjacent 2H protons.

b. Consider the NMR spectra of N:

• The singlet at 1.7 attributes to 3H due to the adjacent carbon with no hydrogen atoms.
• The two singlets at 4.6 attributes to 2H due to the adjacent carbon with no hydrogen atoms.
• The triplet at 2.0 attributes to 2H due to the adjacent 2H protons.
• The multiplet at 1.5 attributes to 2H due to the adjacent 2H and 3H protons.
• The triplet at 0.95 attributes to 3H due to the adjacent 2H protons.

The required structures of M and N are as follows:

Molecular structures of M and N