The two highlighted proton in the given compound has to be compared and which is more acidic has to be determined. Concept Introduction: Acids donate protons, base accepts protons. The strength of the acid can be said by finding the willingness of the acid to give proton. An acid giving away the proton can be represented in the form of equation as shown below, H − A ⇌ − H + A − If the acid “HA” is very much willing to give the proton it holds means then it is known as a strong acid. If it is not very much willing means it is a weak acid. The strength of the acid can be figured out by looking into the conjugate base that is formed on removal of proton. After loss of proton, a negative charge will be created. If the formed negative charge is more stabilized means then the acid is a strong acid and if it is not stabilized means then it is a weak acid. There are few factors which determine the strength of the acid and they are, What atom the charge is present Resonance Induction Orbitals If the charge is on a more electronegative atom, then it is stabilized more. Hence, the compound will be more acidic. If the negative charge is made to participate in resonance, then the negative charge will be stabilized. This increases the stability of the conjugate base and in turn the compound will be more acidic. Pull of the electron density by the more electronegative atom is known as induction. The inductive effects can stabilize or destabilize the conjugate base. If the inductive effect stabilize the conjugate base, then the compound will be acidic. The orbital in which the negative charge is present also plays an important role in stability of the conjugate base. A negative charge on s p hybridized orbital is more stable than the negative charge that is present on s p 3 hybridized orbital. In order to find whether the compound is more acidic or not, the first step is to remove the proton to form conjugate base. Then look for the above four factors.

Question

Want to see the full answer?

Answer 1

Question

Chapter 3.5, Problem 3.19P

Interpretation Introduction

Interpretation:

The two highlighted proton in the given compound has to be compared and which is more acidic has to be determined.

Concept Introduction:

Acids donate protons, base accepts protons. The strength of the acid can be said by finding the willingness of the acid to give proton. An acid giving away the proton can be represented in the form of equation as shown below,

H−A ⇌−H+ A−

If the acid “HA” is very much willing to give the proton it holds means then it is known as a strong acid. If it is not very much willing means it is a weak acid. The strength of the acid can be figured out by looking into the conjugate base that is formed on removal of proton. After loss of proton, a negative charge will be created. If the formed negative charge is more stabilized means then the acid is a strong acid and if it is not stabilized means then it is a weak acid.

There are few factors which determine the strength of the acid and they are,

What atom the charge is present

Resonance

Induction

Orbitals

If the charge is on a more electronegative atom, then it is stabilized more. Hence, the compound will be more acidic.

If the negative charge is made to participate in resonance, then the negative charge will be stabilized. This increases the stability of the conjugate base and in turn the compound will be more acidic.

Pull of the electron density by the more electronegative atom is known as induction. The inductive effects can stabilize or destabilize the conjugate base. If the inductive effect stabilize the conjugate base, then the compound will be acidic.

The orbital in which the negative charge is present also plays an important role in stability of the conjugate base. A negative charge on sp hybridized orbital is more stable than the negative charge that is present on sp3 hybridized orbital.

In order to find whether the compound is more acidic or not, the first step is to remove the proton to form conjugate base. Then look for the above four factors.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Is this aromatic?

CHEM2323 E Tt PS CH03 Draw and name all monobromo derivatives of pentane, C5H11Br. Problem 3-33 Name: Draw structures for the following: (a) 2-Methylheptane (d) 2,4,4-Trimethylheptane Problem 3-35 (b) 4-Ethyl-2,2-dimethylhexane (e) 3,3-Diethyl-2,5-dimethylnonane (c) 4-Ethyl-3,4-dimethyloctane 2 (f) 4-Isopropyl-3-methylheptane KNIE>

Problem 3-42 Consider 2-methylbutane (isopentane). Sighting along the C2-C3 bond: (a) Draw a Newman projection of the most stable conformation. (b) Draw a Newman projection of the least stable conformation. Problem 3-44 Construct a qualitative potential-energy diagram for rotation about the C-C bond of 1,2-dibromoethane. Which conformation would you expect to be most stable? Label the anti and gauche conformations of 1,2- dibromoethane. Problem 3-45 Which conformation of 1,2-dibromoethane (Problem 3-44) would you expect to have the largest dipole moment? The observed dipole moment of 1,2-dibromoethane is µ = 1.0 D. What does this tell you about the actual conformation of the molecule?