Concept explainers
Interpretation:
The common name for the ether whose IUPAC name is ethoxyethane has to be chosen from the given options.
Concept Introduction:
Any organic molecule can be named by using certain rules given by IUPAC (International Union for Pure and applied chemistry). IUPAC name consists of three parts in major namely Prefix suffix and root word.
Prefix represents the substituent present in the molecule and its position in the root name.
Suffix denotes the presence of
Root word represents the longest continuous carbon skeleton of the organic molecule.
Rules for assigning common names to ether:
For obtaining common name for ether, two rules are applicable, one for symmetrical ethers and one for unsymmetrical ethers.
- ✓ For unsymmetrical ethers, the two hydrocarbon groups that is attached to the oxygen atom is arranged in a alphabetical order and the word ether is added. The words are separated by a space. These names have three words with space between them.
- ✓ For symmetrical ethers, prefix di- is used. Then the word ether is added with a space between the two words. These names have two words with space between them.

Want to see the full answer?
Check out a sample textbook solution
Chapter 14 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- Don't used hand raiting and don't used Ai solutionarrow_forward2' P17E.6 The oxidation of NO to NO 2 2 NO(g) + O2(g) → 2NO2(g), proceeds by the following mechanism: NO + NO → N₂O₂ k₁ N2O2 NO NO K = N2O2 + O2 → NO2 + NO₂ Ко Verify that application of the steady-state approximation to the intermediate N2O2 results in the rate law d[NO₂] _ 2kk₁[NO][O₂] = dt k+k₁₂[O₂]arrow_forwardPLEASE ANSWER BOTH i) and ii) !!!!arrow_forward
- E17E.2(a) The following mechanism has been proposed for the decomposition of ozone in the atmosphere: 03 → 0₂+0 k₁ O₁₂+0 → 03 K →> 2 k₁ Show that if the third step is rate limiting, then the rate law for the decomposition of O3 is second-order in O3 and of order −1 in O̟.arrow_forward10.arrow_forwardDon't used Ai solution and don't used hand raitingarrow_forward
- Organic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning

