Consider the reaction of (R)-2-chloro-3-methylbutane with sodium iodide to form a product.

1(a) Draw the reaction scheme with the correct stereochemistry (reactant + NaI → product

+ NaCl). Circle the nucleophile and draw a rectangle around the electrophile.

1(b) What is the symbol used for mechanism shown in 1(a)?

1(c) If the sodium iodide was replaced with sodium hydroxide, the product is an ALKENE. Draw a reaction MECHANISM to show how this happens.

1(d) Draw the reaction energy diagram for the reaction in 1(c) and label the activation

energy.

1(e) Using any alcohol with five carbons, and any carboxylic acid with six carbons, draw a

reaction to show how we would make an ester.

1(f) Describe the practical on esters.

See explanation and image attached

Explanation:

Now let us consider the reaction closely, The reaction of (R)-2-chloro-3-methylbutane with sodium iodide is an SN2 reaction that occurs with inversion of stereochemistry.

When the OH^- is used, the reaction proceeds by E2 elimination giving the alkene product as shown in the image attached to this answer.

The reaction of a five-carbon alkanol and an acid to form an ester is also shown in the image attached to this answer.

To prepare an ester. the glassware must first be dried. The alkanol and alkanoic acid are added and refluxed in a magnetic stirrer and conc H2SO4 is added. A thermometer should be present in the set up. The refluxing should continue for about 60 minutes.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

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