The reason for the fact that HF is a weak acid, whereas HCl , HBr and HI is all strong acids is to be stated. Concept introduction: Strong acids give away H + ion easily, whereas weak acids do not. The acidic strength of hydrogen halides ( H − X ) depends on their bond strength. Higher the bond strength, lower will be the acidic strength. To determine: An explanation for the fact that HF is a weak acid, whereas HCl , HBr and HI is all strong acids.

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(a) The following synthesis of the molecule shown in the circle has a major problem. What is this problem? (2 pts) 1) HBr (no peroxides) 2) H- NaNH2 Br 3) NaNH, 4) CH3Br 5) H2, Pd (b) Starting with the molecule shown below and any other materials with two carbons or less, write out an alternate synthesis of the circled molecule. More than one step is needed. Indicate the reagent(s) and the major product in all the steps in your synthesis. (5 pts) 2024 Fall Term (1) Organic Chemistry 1 (Lec) CHEM 22204 02[6386] (Hunter College) (c) Using the same starting material as in part (b) and any other materials win two carpons or less, write out syntheses of the circled molecules shown below. More than one step is needed in each case. Indicate the reagent(s) and the major product in all the steps in your synthesis. You may use reactions and products from your synthesis in part (b). (5 pts)

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Chapter 20, Problem 66E

Interpretation Introduction

Interpretation: The reason for the fact that HF is a weak acid, whereas HCl,HBr and HI is all strong acids is to be stated.

Concept introduction: Strong acids give away H+ ion easily, whereas weak acids do not. The acidic strength of hydrogen halides (H−X) depends on their bond strength. Higher the bond strength, lower will be the acidic strength.

To determine: An explanation for the fact that HF is a weak acid, whereas HCl,HBr and HI is all strong acids.

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(a) The following synthesis of the molecule shown in the circle has a major problem. What is this problem? (2 pts) 1) HBr (no peroxides) 2) H- NaNH2 Br 3) NaNH, 4) CH3Br 5) H2, Pd (b) Starting with the molecule shown below and any other materials with two carbons or less, write out an alternate synthesis of the circled molecule. More than one step is needed. Indicate the reagent(s) and the major product in all the steps in your synthesis. (5 pts) 2024 Fall Term (1) Organic Chemistry 1 (Lec) CHEM 22204 02[6386] (Hunter College) (c) Using the same starting material as in part (b) and any other materials win two carpons or less, write out syntheses of the circled molecules shown below. More than one step is needed in each case. Indicate the reagent(s) and the major product in all the steps in your synthesis. You may use reactions and products from your synthesis in part (b). (5 pts)

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