When the regex matches the first time (on "A B"), this part of the string in consumed by the engine, so it is not matched again, even though your regex has the global ('g') flag.

You could achieve the expected result by using a positive lookahead ((?=PATTERN)) instead, that won't consume the match:

value = "Hello I B M"
value = value.replace(/([A-Z])\s(?=[A-Z])/g, '$1')
console.log(value) // Prints "Hello IBM"

To make it not remove the space if the next uppercase letter is the first in a word, you can increment the lookahead pattern with using a word boundary \b to make that restriction:

value = "Hello I B M Dude"
value = value.replace(/([A-Z])\s(?=[A-Z]\b)/g, '$1')
console.log(value) // Prints "Hello IBM Dude"

Note: As @CasimirHyppolite noted, the following letter has to be made optional, or the second regex won't work if the last character of the string is uppercase. Thus, the pattern ([^A-Za-z]|$), which can be read as "not a letter, or the end of the string".

Edit: Simplify lookahead from (?=A-Z) to (?=[A-Z]\b) as suggested by @hwnd

This removes 2 or more spaces only inside <p class="text_obisnuit"> and </p> and keep any other multiple spaces.

• Ctrl+H
• Find what: (?:<p class="text_obisnuit">|\G)(?:(?!</p>).)*?\s\K\s+
• Replace with: LEAVE EMPTY
• check Wrap around
• check Regular expression
• DO NOT CHECK . matches newline depending if you want to match multiple lines or not.
• Replace all

Explanation:

(?:                         # start non capture group
  <p class="text_obisnuit"> # literally
 |                          # OR
  \G                        # restart from position of last match
)                           # end group
(?:                         # start non capture group
  (?!</p>)                  # negative lookahead, make sure we haven't reach </p>
  .                         # any character
)*?                         # group may appear 0 or more times, not greedy
\s                          # a space
\K                          # forget all we have seen until this position
\s+                         # 1 or more spaces

Given text:

other     text

<p class="text_obisnuit">  The context of articles,   stories, and conversations helps you     figure out and understand the meaning   of English words in the text that are new to you.   </p>

other    text

Result for given example:

other     text

<p class="text_obisnuit"> The context of articles, stories, and conversations helps you figure out and understand the meaning of English words in the text that are new to you. </p>

other    text

Note: it keeps space just after <p...> and just before </p>

If you want to remove these spaces, you have to run another regex:

• Ctrl+H
• Find what: (?<=<p class="text_obisnuit">)\s+|\s+(?=</p>)
• Replace with: LEAVE EMPTY
• UNcheck Match case
• check Wrap around
• check Regular expression
• Replace all

Explanation:

(?<=                        # start positive lookbehind, make sure we have 
  <p class="text_obisnuit"> # literally
)                           # end lookbehind
\s+                         # 1 or more spaces
|                           # OR
\s+                         # 1 or more spaces
(?=                         # start positive lookahead
  </p>                      # literally
)                           # end lookahead

Result for given example:

other     text

<p class="text_obisnuit">The context of articles, stories, and conversations helps you figure out and understand the meaning of English words in the text that are new to you.</p>

other    text

Given that you also want to cover tabs, newlines, etc, just replace \s\s+ with ' ':

string = string.replace(/\s\s+/g, ' ');

If you really want to cover only spaces (and thus not tabs, newlines, etc), do so:

string = string.replace(/  +/g, ' ');