Comparing to diphenylamine ( pK a = 0 .79 ) , aniline ( conjugate acid pK a = 4 .63 ) is a stronger base. Reason for this difference has to be marked. Concept introduction: Amines are the derivatives of ammonia ( N H 3 ) with one or more of the hydrogens replaced with a substituent such as an alkyl group. Depending on the number of carbon side chain of the nitrogen, different types of amines can form. If nitrogen atom of amine group is attached only to alkyl groups, then it is an aliphatic amine. If nitrogen atom of amine group is attached to at least one aryl groups, then it is an aromatic amine. Lone pair present on nitrogen of aromatic amine takes part in resonance by interacting with the π system of the aromatic ring. This stabilizing interaction will be lost when the aromatic amine is protonated. Therefore, aliphatic amine is more basic than comparing to an aromatic amine.

Question

1) The isoamyl acetate report requires eight paragraphs - four for comparison of isoamyl alcohol and isoamyl acetate (one paragraph each devoted to MS, HNMR, CNMR and IR) and four for comparison of acetic acid and isoamyl acetate ((one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of…

Answer 1

Question

Chapter 23, Problem 23.29P

Interpretation Introduction

Interpretation:

Comparing to diphenylamine (pKa= 0.79), aniline (conjugate acid pKa = 4.63) is a stronger base. Reason for this difference has to be marked.

Concept introduction:

Amines are the derivatives of ammonia (NH3) with one or more of the hydrogens replaced with a substituent such as an alkyl group.

Depending on the number of carbon side chain of the nitrogen, different types of amines can form.

If nitrogen atom of amine group is attached only to alkyl groups, then it is an aliphatic amine.

If nitrogen atom of amine group is attached to at least one aryl groups, then it is an aromatic amine.

Lone pair present on nitrogen of aromatic amine takes part in resonance by interacting with the π system of the aromatic ring. This stabilizing interaction will be lost when the aromatic amine is protonated. Therefore, aliphatic amine is more basic than comparing to an aromatic amine.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

1) The isoamyl acetate report requires eight paragraphs - four for comparison of isoamyl alcohol and isoamyl acetate (one paragraph each devoted to MS, HNMR, CNMR and IR) and four for comparison of acetic acid and isoamyl acetate ((one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of…

№3 Fill in the below boxes. HN 1. LAH 2. H3O+ NH2

For the photochemical halogenation reaction below, draw both propagation steps and include the mechanism arrows for each step. H CH ot CH3 CI-CI MM hv of CH H-CI CH3 2nd attempt See Periodic Table See Hint Draw only radical electrons; do not add lone pair electrons. Note that arrows cannot meet in "space," and must end at either bonds or at atoms. 1 i Add the missing curved arrow notation to this propagation step. 20 H ن S F P H CI Br 品