Question

For the reaction below, focus only on 2-methyl-2-butanol (starting material) and 2-chloro-2-methylbutane (desired product). We will not analyze HCl or H₂O by IR.

1. From the two IR spectra below, identify which spectrum belongs to which molecule. Then label the IR spectrum with the molecule’s characteristic IR peaks. Highlight or place a box around the most characteristic peak in the starting material’s IR. You do not need to label the fingerprint region.

For the reaction below, focus only on 2-methyl-2-butanol (starting material) and 2-chloro-2-methylbutane (desired product). We will not analyze HCl or H,O by IR. OH + H-CI + HO 2-methyl-2-butanol 2-chloro-2-methylbutane a) From the two IR spectra below, identify which spectrum belongs to which molecule. Then label the IR spectrum with the molecule's characteristic IR peaks. Highlight or place a box around the most characteristic peak in the starting material's IR. You do not need to label the fingerprint region. IR spectrum of: mm 000 20.00 ISRO 500 RAVENUMBERS IR spectrum of: 0.81 0.6 Relative Transmittance 0.4 0.2 3000 1000 2000 Wavenumber (cm-1) b) Using IR as your only form of characterization, how would you know that the reaction is complete? Explain.

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Answer 1

Answer::::

Here in given reaction transformation starting material is an tertary alcohol whene react with an acid HCl first give protionation an form carbocation with removal of H2o, then Cl- get attack on formed carbocation to get desired final product via SN1 reaction Mechanisum.

Major peak in Given IR spectrum of Starting material that is tertary alcohol :

1) O-H broad straching frequency observed at 3300 - 3500 cm-1, witch matches to our compound.

2) C-H assymetric straching frequency observed at Sp3 hybridized carbon hydrogen at 2900 cm-1.

these are two main Functional group region peacks are observed in given ir spectrum

Major peak in Given IR spectrum of Final Product :

1) 0nly C-H sp3 hybridized straching frequency observed at 2900-3000 cm-1.

According to Analysing both spectrum here in final product there are absent of O-H broad band hence our reaction is completed.