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nikklg [1K]
3 years ago
12

Give the product expected when the following alcohol reacts with pyridinium chlorochromate (PCC). (Assume that PCC is present in

excess.) The starting material is a 4 carbon ring where carbon 1 is bonded to O H and C H 3. Carbon 3 is bonded to C H 2 C H 2 O H. This reacts with P C C in C H 2 C L 2 to give the product. Draw the product.
Chemistry
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

Kindly check the explanation section.

Explanation:

Based on the description of the reacting -OH group containing Compound, the drawing of the chemical compound is given in the attached picture.

So, without mincing words let's dive straight into the solution to the question.

The reaction between the OH containing compound and PCC is an oxidation reaction.

Looking at the carbon number 1 which the first OH group and CH3 are attached to. Oxidation can not occur here as tertiary alcohol can not be oxidize.

Hence, the second OH will be oxidized into a carbonyl group, C = O. Kindly note that when alcohol oxidizes it turns into an aldehyde.

The equation for the reaction is also given the the attached picture.

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How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?​
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3.72 mol Hg

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
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Explanation:

<u>Step 1: Define</u>

D = 13.6 g/mL

54.8 mL Hg

<u>Step 2: Identify Conversions</u>

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Find</u>

13.6 g/mL = x g / 54.8 mL

x = 745.28 g Hg

<u>Step 4: Convert</u>

<u />745.28 \ g \ Hg(\frac{1 \ mol \ Hg}{200.59 \ g \ Hg} ) = 3.71544 mol Hg

<u>Step 5: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

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8 0
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